Water is filled in a flask up to a height of 50cm. The bottom of the flask is circular with radius 10cm. If the atmospheric pressure is $1.01\times {10}^5$ Pa. The force exerted by the water on the bottom is (Take $g=10m/s^2$ and density of water = 1000 kg / $m^3$)

The pressure at the surface of the water is equal to the atmospheric pressure $p_0$. The pressure at the bottom is  $p=p_0+hdg$

$$\begin{gathered} p=1.01\times {10}^{5}Pa+\left(0.50m\right)(1000kg/m^3)(10m/s^2) \\ p=1.01\times {10}^{5}Pa+0.05\times {10}^{5}Pa \\ p=1.06\times {10}^{5}Pa \end{gathered}$$

The area of the bottom  $=\pi r^2$
$$\begin{gathered} =314\times {10}^{-3} \\ =\frac{314}{1000}=0.314 \end{gathered}$$                   
The force on the bottom is, therefore,
$$\begin{gathered} F=P\times \pi r^2 \\ F=(1.06\times {10}^{5}Pa)\left(0.0314m^2\right) \\ F=3328N \end{gathered}$$